892. Surface Area of 3D Shapes(Leetcode每日一题-2020.03.25)

Problem

On a N * N grid, we place some 1 * 1 * 1 cubes.

Each value v = grid[i][j] represents a tower of v cubes placed on top of grid cell (i, j).

Return the total surface area of the resulting shapes.

Example1

Input: [[2]]
Output: 10

Example2

Input: [[1,2],[3,4]]
Output: 34

Example3

Input: [[1,0],[0,2]]
Output: 16

Example4

Input: [[1,1,1],[1,0,1],[1,1,1]]
Output: 32

Example5

Input: [[2,2,2],[2,1,2],[2,2,2]]
Output: 46

Solution

对于每一个grid[i][j] ，如果grid[i][j]大于零，那么其表面积就是2 + 4 * grid[i][j]。
但是grid[i-1][j]和grid[i][j-1]，即左侧和上侧会造成遮挡，要减去遮挡面积 min(grid[i][j],grid[i-1][j]) *2 和min(grid[i][j],grid[i][j-1]) *2。

class Solution {
public:
    int surfaceArea(vector<vector<int>>& grid) {
        if(grid.empty())
            return 0;
        int r = grid.size();
        int c = grid[0].size();

        int surface_area = 0;
        for(int i = 0;i<r;++i)
        {
            for(int j = 0;j<c;++j)
            {
                int cubes = grid[i][j];
                if(cubes > 0)
                {
                    surface_area += (2 + cubes * 4);

                    if(i - 1 >= 0)
                    {
                        surface_area -= (min(cubes,grid[i-1][j]) * 2);
                    }

                    if(j - 1 >= 0)
                    {
                        surface_area -= (min(cubes,grid[i][j-1]) * 2);
                    }
                }
            }
        }

        return surface_area;

    }
};

Ref

https://leetcode-cn.com/problems/surface-area-of-3d-shapes/solution/shi-li-you-tu-you-zhen-xiang-jiang-jie-yi-kan-jiu-/

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